The perimeter of a square is increasing at a rate of $5$ meters per hour. At a certain instant, the perimeter is $30$ meters. What is the rate of change of the area of the square at that instant (in square meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{75}{4}$ (Choice B) B $\dfrac{5}{4}$ (Choice C) C $25$ (Choice D) D $\dfrac{25}{16}$
Answer: Setting up the math Let... $s(t)$ denote the square's side at time $t$, $A(t)$ denote the square's area at time $t$, and $P(t)$ denote the square's perimeter at time $t$. We are given that $P'(t)=5$, We are also given that $P(t_0)=30$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures $P(t)$ and $s(t)$ relate to each other through the formula for the perimeter of a square: $P(t)=4s(t)$ We can differentiate both sides to find an expression for $P'(t)$ : $P'(t)=4s'(t)$ $A(t)$ and $s(t)$ relate to each other through the formula for the area of a square: $A(t)=[s(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2s(t)s'(t)$ Using the information to solve Let's plug ${P(t_0)}={30}$ into the expression for $P(t_0)$ : $\begin{aligned} {P(t_0)}&=4s(t_0) \\\\ {30}&=4s(t_0) \\\\ {\dfrac{15}{2}}&={s(t_0)} \end{aligned}$ Let's plug ${P'(t_0)}={5}$ into the expression for $P'(t_0)$ : $\begin{aligned} {P'(t_0)}&=4s'(t_0) \\\\ {5}&=4s'(t_0) \\\\ C{\dfrac{5}{4}}&=C{s'(t_0)} \end{aligned}$ Now let's plug ${s(t_0)}={\dfrac{15}{2}}$ and $C{s'(t_0)}=C{\dfrac{5}{4}}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2{s(t_0)}C{s'(t_0)} \\\\ &=2\left({\dfrac{15}{2}}\right)\left(C{\dfrac{5}{4}}\right) \\\\ &=\dfrac{75}{4} \end{aligned}$ In conclusion, the rate of change of the area of the square at that instant is $\dfrac{75}{4}$ square meters per hour. Since the rate of change is positive, we know that the area is increasing.